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Question

The line 2x + 3y = 12 meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to AB meets the x-axis and the line AB at C and E respectively. If O is the origin of coordinates, find the area of figure OCEB.


Solution

The given line is 2x + 3y = 12, which can be written as

x6+y4=1              ... (1)

So, the coordinates of the points A and B are (6, 0) and (0, 4), respectively.



The equation of the line perpendicular to line (1) is

x4-y6+λ=0

This line passes through the point (5, 5).

54-56+λ=0λ=-512

Now, substituting the value of λ in x4-y6+λ=0, we get:
x4-y6-512=0x53-y52=1            ...(2)

        
Thus, the coordinates of intersection of line (1) with the x-axis is C 53,0.

To find the coordinates of E, let us write down equations (1) and (2) in the following manner:

2x+3y-12=0        ... (3)

3x-2y-5=0          ... (4)

Solving (3) and (4) by cross multiplication, we get:

x-15-24=y-36+10=1-4-9x=3, y=2

Thus, the coordinates of E are (3, 2).

From the figure,

EC=53-32+0-22=2133



EA=6-32+0-22=13

Now,

Area OCEB=Area OAB-Area CAEArea OCEB=12×6×4-12×2133×13                             =233 sq units

Mathematics
RD Sharma XI (2015)
Standard XI

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